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Chapter 16

Solubility and �Complex Ion Equilibria

Section 16.1

Solubility Equilibria and the Solubility Product

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(16.1) Solubility equilibria and the solubility product
(16.2) Precipitation and qualitative analysis
(16.3) Equilibria involving complex ions

Chapter 16

Table of Contents

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Equilibria

When a typical ionic solid dissolves in water, it separates into cations and anions
Example

Ions formed - Ca²⁺ and F^–
In this reaction, when solid salt is first added, no ions are present
As dissolution proceeds, the ionic concentration increases

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Section 16.1

Solubility Equilibria and the Solubility Product

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Equilibria (Continued)

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Section 16.1

Solubility Equilibria and the Solubility Product

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Solubility Equilibria

Solubility product constant (K_sp)

Equilibrium expression constant that represents the dissolution of an ionic solid in water
Known as solubility product

Solubility equilibrium is unaffected by:

Excess solid formed
Size of particles present

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Section 16.1

Solubility Equilibria and the Solubility Product

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Solubility Equilibria (Continued)

Differences between the solubility of a given solid and its solubility product

Solubility is an equilibrium position
Solubility product is an equilibrium constant

Has only one value for a given solid at a given temperature

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Section 16.1

Solubility Equilibria and the Solubility Product

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Interactive Example 16.2 - Calculating K_sp from Solubility II

Calculate the K_sp value for bismuth sulfide (Bi₂S₃), which has a solubility of 1.0×10^–15 mol/L at 25°C

Section 16.1

Solubility Equilibria and the Solubility Product

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Interactive Example 16.2 - Solution

Section 16.1

Solubility Equilibria and the Solubility Product

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Interactive Example 16.2 - Solution (Continued 1)

Thus, the equilibrium concentrations of these ions will be determined by the amount of salt that dissolves to reach equilibrium, which in this case is 1.0×10^–15 mol/L
Since each Bi₂S₃ unit contains 2Bi³⁺ and 3S^2– ions,

Section 16.1

Solubility Equilibria and the Solubility Product

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Interactive Example 16.2 - Solution (Continued 2)

The equilibrium concentrations are

Therefore,

Section 16.1

Solubility Equilibria and the Solubility Product

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Exercise

Approximately 0.14 g nickel(II) hydroxide, Ni(OH)₂(s), dissolves per liter of water at 20°C

Calculate K_sp for Ni(OH)₂(s) at this temperature

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K_sp = 1.4×10^–8

Section 16.1

Solubility Equilibria and the Solubility Product

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Interactive Example 16.3 - Calculating Solubility from K_sp

The K_sp value for copper(II) iodate, Cu(IO₃)₂, is 1.4×10^–7 at 25°C

Calculate its solubility at 25°C

Section 16.1

Solubility Equilibria and the Solubility Product

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Interactive Example 16.3 - Solution

Section 16.1

Solubility Equilibria and the Solubility Product

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Interactive Example 16.3 - Solution (Continued 1)

We do this in the usual way by specifying the initial concentrations (before any solid has dissolved) and then defining the change required to reach equilibrium
Since in this case we do not know the solubility, we will assume that x mol/L of the solid dissolves to reach equilibrium
The 1:2 stoichiometry of the salt means that

Section 16.1

Solubility Equilibria and the Solubility Product

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Interactive Example 16.3 - Solution (Continued 2)

The concentrations are as follows:

Section 16.1

Solubility Equilibria and the Solubility Product

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Interactive Example 16.3 - Solution (Continued 3)

Substituting the equilibrium concentrations into the expression for K_sp gives

Thus, the solubility of solid Cu(IO₃)₂ is 3.3×10^–3 mol/L

Section 16.1

Solubility Equilibria and the Solubility Product

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Exercise

Calculate the solubility of each of the following compounds in moles per liter

Ag₃PO₄K_sp = 1.8×10^–18

CaCO₃ K_sp = 8.7×10^–9

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1.6×10^–5 mol/L

9.3×10^–5mol/L

Section 16.1

Solubility Equilibria and the Solubility Product

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Relative Solubilities

One must be careful in using K_sp values to predict the relative solubilities of a group of salts
Possible cases

Salts being compared produce the same number of ions

K_sp values cannot be compared to determine relative solubilities

Salts being compared produce varying number of ions

Cannot predict relative solubilities using K_sp values

Section 16.1

Solubility Equilibria and the Solubility Product

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Common Ion Effect

The solubility of a solid becomes low when the solution already contains ions common to the solid

A potassium chromate solution being added to aqueous silver nitrate, forming silver chromate

Section 16.1

Solubility Equilibria and the Solubility Product

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Critical Thinking

What if all you know about two salts is that the value of K_sp for salt A is greater than that of salt B?

Why can we not compare relative solubilities of the salts?
Use numbers to show how salt A could be more soluble than salt B, and how salt B could be more soluble than salt A

Section 16.1

Solubility Equilibria and the Solubility Product

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Interactive Example 16.4 - Solubility and Common Ions

Calculate the solubility of solid CaF₂ (K_sp = 4.0×10^–11) in a 0.025-M NaF solution

Section 16.1

Solubility Equilibria and the Solubility Product

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Interactive Example 16.4 - Solution

Section 16.1

Solubility Equilibria and the Solubility Product

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Interactive Example 16.4 - Solution (Continued 1)

Substituting the equilibrium concentrations into the expression for K_sp gives

Section 16.1

Solubility Equilibria and the Solubility Product

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Interactive Example 16.4 - Solution (Continued 2)

Assuming that 2x is negligible compared with 0.025 (since K_sp is small) gives

The approximation is valid (by the 5% rule)

Thus 6.4 ×10^–8 mole of solid CaF₂ dissolves per liter of the 0.025-M NaF solution

Section 16.1

Solubility Equilibria and the Solubility Product

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pH and Solubility

Increase in pH

Decreases solubility
Forces the equilibrium to the left

Decrease in pH

Increases solubility
Equilibrium shifts to the right
If the anion X^– is an effective base, the salt MX will show increased solubility in an acidic solution

Section 16.1

Solubility Equilibria and the Solubility Product

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pH and Solubility (Continued)

Exception - AgCl has the same solubility in acid as in pure water

Cl^– ion is a weak base

No HCl molecules are formed

Adding H⁺ ions to a solution that contains Cl^– ions does not affect:

Concentration of the Cl^– ion
Solubility of the chloride salt

Section 16.1

Solubility Equilibria and the Solubility Product

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Critical Thinking

You and a friend are studying for a chemistry exam

What if your friend tells you that since acids are very reactive, all salts are more soluble in aqueous solutions of acids than in water?
How would you explain to your friend that this is not true?
Use a specific example to defend your answer

Section 16.1

Solubility Equilibria and the Solubility Product

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Precipitation

Ion product (Q)

Defined similar to the expression for K_sp for a given solid
Exception - Initial concentrations are used instead of equilibrium concentrations

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Section 16.2

Precipitation and Qualitative Analysis

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Relationship between Q and K_sp

One can predict the possibility of precipitation by considering the relationship between Q and K_sp

Q > K_sp

Precipitation occurs and will continue until the concentrations are reduced to the point that they satisfy K_sp

Q < K_sp

No precipitation occurs

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Section 16.2

Precipitation and Qualitative Analysis

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Interactive Example 16.5 - Determining Precipitation Conditions

A solution is prepared by adding 750.0 mL of 4.00×10^–3 M Ce(NO₃)₃ to 300.0 mL of 2.00×10^–2 M KIO₃

Will Ce(IO₃)₃ (K_sp = 1.9×10^–10) precipitate from this solution?

Section 16.2

Precipitation and Qualitative Analysis

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Interactive Example 16.5 - Solution

First, we calculate [Ce³⁺]₀ and [IO₃^–]₀ in the mixed solution before any reaction occurs:

Section 16.2

Precipitation and Qualitative Analysis

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Interactive Example 16.5 - Solution (Continued)

The ion product for Ce(IO₃)₃ is

Since Q is greater than K_sp, Ce(IO₃)₃ will precipitate from the mixed solution

Section 16.2

Precipitation and Qualitative Analysis

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Calculating Equilibrium Concentrations after Precipitation

Step 1 - Determine if the product is formed when the solutions are mixed

Calculate the concentration of the ions in the mixed solution to determine Q

Step 2 - Run the reaction to completion

Note - If a reaction virtually goes to completion when two solutions are mixed, it is necessary to conduct stoichiometric calculations prior to equilibrium calculations

Section 16.2

Precipitation and Qualitative Analysis

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Calculating Equilibrium Concentrations after Precipitation Occurs (Continued)

Step 3 - Allow the system to adjust to equilibrium and determine the concentrations of ions in the solution
Step 4 - Substitute the expressions derived in step 3 to determine K_sp

Section 16.2

Precipitation and Qualitative Analysis

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Interactive Example 16.6 - Precipitation

A solution is prepared by mixing 150.0 mL of 1.00×10^–2 M Mg(NO₃)₂ and 250.0 mL of 1.00×10^–1 M NaF

Calculate the concentrations of Mg²⁺ and F^– at equilibrium with solid MgF₂ (K_sp = 6.4×10^–9)

Section 16.2

Precipitation and Qualitative Analysis

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Interactive Example 16.6 - Solution

First, determine whether solid MgF₂ forms

To do this, we need to calculate the concentrations of Mg²⁺ and F^– in the mixed solution and find Q

Section 16.2

Precipitation and Qualitative Analysis

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Interactive Example 16.6 - Solution (Continued 1)

Since Q is greater than K_sp, solid MgF₂ will form

The next step is to run the precipitation reaction to completion

Section 16.2

Precipitation and Qualitative Analysis

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Interactive Example 16.6 - Solution (Continued 2)

Note that excess F^– remains after the precipitation reaction goes to completion

The concentration is

Although we have assumed that Mg²⁺ is completely consumed, we know that [Mg²⁺] will not be zero at equilibrium

Section 16.2

Precipitation and Qualitative Analysis

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Interactive Example 16.6 - Solution (Continued 3)

Section 16.2

Precipitation and Qualitative Analysis

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Interactive Example 16.6 - Solution (Continued 4)

Section 16.2

Precipitation and Qualitative Analysis

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Selective Precipitation

Method used to separate mixtures of metal ions in aqueous solution
Involves using a reagent whose anion forms a precipitate with only one or a few of the metal ions in the mixture

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Section 16.2

Precipitation and Qualitative Analysis

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Selective Precipitation - Example

Consider a solution containing Ba²⁺ and Ag⁺ ions

NaCl is added to this solution
Result

AgCl precipitates as a white solid
BaCl₂ is soluble, so Ba²⁺ ions remain in the solution

Section 16.2

Precipitation and Qualitative Analysis

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Example 16.7 - Selective Precipitation

A solution contains 1.0×10^–4 M Cu⁺ and 2.0×10^–3 M Pb²⁺

If a source of I^–is added gradually to this solution, will PbI₂ (K_sp = 1.4×10^–8) or CuI (K_sp= 5.3×10^–12) precipitate first?
Specify the concentration of I^– necessary to begin precipitation of each salt

Section 16.2

Precipitation and Qualitative Analysis

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Example 16.7 - Solution

For PbI₂, the K_sp expression is

Since [Pb²⁺] in this solution is known to be 2.0×10^–3 M, the greatest concentration of I^–that can be present without causing precipitation of PbI₂ can be calculated from the K_sp expression

Section 16.2

Precipitation and Qualitative Analysis

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Example 16.7 - Solution (Continued 1)

Any I^– in excess of this concentration will cause solid PbI₂ to form
Similarly, for CuI, the K_sp expression is

Section 16.2

Precipitation and Qualitative Analysis

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Example 16.7 - Solution (Continued 2)

A concentration of I^– in excess of 5.3×10^–8 M will cause formation of solid CuI
As I^– is added to the mixed solution, CuI will precipitate first, since the [I^–] required is less

Therefore, Cu⁺ would be separated from Pb²⁺ using this reagent

Section 16.2

Precipitation and Qualitative Analysis

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Exercise

A solution is 1×10^–4 M in NaF, Na₂S, and Na₃PO₄

What would be the order of precipitation as a source of Pb²⁺ is added gradually to the solution?
The relevant K_sp values are:

K_sp(PbF₂) = 4 × 10^–8

K_sp(PbS) = 7 × 10^–29

K_sp[Pb₃(PO₄)₂] = 1 × 10^–54

PbS(s) will form first, followed by Pb₃(PO₄)₂(s), and PbF₂(s) will form last

Section 16.2

Precipitation and Qualitative Analysis

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Selective Precipitation of Metal Sulfide Salts

Metal sulfide salts differ in solubilities
Sulfide ion is used as a precipitating reagent to separate metal ions

Advantage - Sulfide ion is basic in nature

Its concentration can be controlled by regulating the pH of the solution

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Section 16.2

Precipitation and Qualitative Analysis

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Qualitative Analysis

Involves separating common cations into five major groups based on solubilities

Individual ions are then identified from each of the groups
Group I - Insoluble chlorides
Group II - Sulfides insoluble in acid solution
Group III - Sulfides insoluble in basic solution
Group IV- Insoluble carbonates
Group V - Alkali metal and ammonium ions

Section 16.2

Precipitation and Qualitative Analysis

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Group I - Insoluble Chlorides

After adding dilute HCl to a solution that contains a mixture of common cations:

Ag⁺, Pb⁺, and Hg²⁺ precipitate as insoluble chlorides
Other chlorides are soluble and remain in the solution

Group I precipitate is removed

The remaining ions are treated with sulfide ion

Section 16.2

Precipitation and Qualitative Analysis

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Group II - Sulfides Insoluble in Acid Solution

The solution remains acidic after removing insoluble chlorides due to the presence of HCl

When H₂S is added to the solution:

Highly insoluble sulfides will precipitate due to the low concentration of S^2–
Soluble sulfides will remain dissolved and the precipitate of the insoluble salt is removed

Section 16.2

Precipitation and Qualitative Analysis

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Group III - Sulfides Insoluble in Basic Solution

The solution is basic at this stage
More H₂S is added
Cations that precipitate as sulfides include Co²⁺, Zn²⁺, Mn²⁺, Ni²⁺, and Fe²⁺

Cr³⁺ and Al³⁺ ions precipitate as insoluble hydroxides

Precipitate is separated from the solution

Section 16.2

Precipitation and Qualitative Analysis

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Group IV - Insoluble Carbonates

All ions, except those from Groups 1A and 2A of the periodic table, have precipitated in this stage
Group 2A cations form insoluble carbonates

Can be precipitated by the addition of CO₃^2– to the solution

Section 16.2

Precipitation and Qualitative Analysis

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Group V - Alkali Metal and Ammonium Ions

Ions remaining in the solution at this stage

Group 1A cations
NH₄⁺ ion

Ions form soluble salts with common anions

Section 16.2

Precipitation and Qualitative Analysis

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Complex Ion Equilibria

Complex ion: Charged species consisting of a metal ion surrounded by ligands

Ligand - Lewis base

Molecule or ion that has a lone pair that can be donated to an empty orbital on the metal ion to form a covalent bond
Coordination number - Number of ligands attached to a metal ion

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Section 16.3

Equilibria Involving Complex Ions

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Formation (Stability) Constant

Equilibrium constant for each step of the formation of a complex ion by the addition of an individual ligand to a metal ion or complex ion in aqueous solution

Usually, the total concentration of the ligand is larger than the total concentration of the metal ion

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Section 16.3

Equilibria Involving Complex Ions

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Interactive Example 16.8 - Complex Ions

Section 16.3

Equilibria Involving Complex Ions

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Interactive Example 16.8 - Solution

The concentrations of the ligand and metal ion in the mixed solution before any reaction occurs are

Section 16.3

Equilibria Involving Complex Ions

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Interactive Example 16.8 - Solution (Continued 1)

Since [S₂O₃^–2]₀ >> [Ag⁺]₀, and since K₁ and K₂ are large, both formation reactions can be assumed to go to completion, and the net reaction in the solution is as follows:

Section 16.3

Equilibria Involving Complex Ions

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Interactive Example 16.8 - Solution (Continued 2)

Note that Ag⁺ is limiting and that the amount of S₂O₃^2– consumed is negligible

Also note that since all these species are in the same solution, the molarities can be used to do the stoichiometry problem

Of course, the concentration of Ag⁺ is not zero at equilibrium, and there is some Ag(S₂O₃)^– in the solution

To calculate the concentrations of these species, we must use the K₁ and K₂ expressions

Section 16.3

Equilibria Involving Complex Ions

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Interactive Example 16.8 - Solution (Continued 3)

We can calculate the concentration of Ag(S₂O₃)^– from K₂

We can calculate [Ag⁺] from K₁

Section 16.3

Equilibria Involving Complex Ions

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Interactive Example 16.8 - Solution (Continued 4)

These results show that [Ag(S₂O₃)₂^3–] >> [Ag(S₂O₃)^–] >> [Ag⁺]

Thus, the assumption is valid that essentially all the original Ag⁺ is converted to Ag(S₂O₃)₂^3– at equilibrium

Section 16.3

Equilibria Involving Complex Ions

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Complex Ions and Solubility

Ionic solids that are nearly water-insoluble must be dissolved somehow in aqueous solutions

Example - When the various qualitative analysis groups are precipitated out, the precipitates must be redissolved to separate the ions within each group

Section 16.3

Equilibria Involving Complex Ions

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Strategies for Dissolving a Water-Insoluble Ionic Solid

If the anion of the solid is a good base, the solubility is greatly increased by acidifying the solution
When the anion is not sufficiently basic, the ionic solid can be dissolved in a solution containing a ligand that forms stable complex ions with its cation

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Section 16.3

Equilibria Involving Complex Ions

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Strategies for Dissolving a Water-Insoluble Ionic Solid (Continued)

Some insoluble solids have to go through a combination of reactions to be dissolved
The solubility of many salts increases with temperature

Simple heating is enough to make a salt sufficiently soluble

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Section 16.3

Equilibria Involving Complex Ions